Class 11 RD Sharma Solutions - Chapter 29 Limits - Exercise 29.6 | Set 2

Question 14.  Lim_n→∞{1² + 2² + ................ + n²}/(n³)

Solution:

We have,

Lim_n→∞{1² + 2² + ................ + n²}/(n³)

= Lim_n→∞[n(n+1)(2n+1)]/6n³

= Lim_n→∞[(n+1)(2n+1)]/6n²

=

When n → ∞, (1/n) → 0

= 2/6

= 1/3

Question 15. Lim_n→∞{1 + 2 + 3 + 4 +................ + n - 1}/n²

Solution:

We have,

Lim_n→∞{1 + 2 + 3 + 4 +................ + n - 1}/n²

= Lim_n→∞[n(n - 1)/2n²]

= Lim_n→∞[n² - n/2n²]

= Lim_n→∞(1/2 - 1/2n)

When n → ∞, (1/n) → 0

= 1/2

Question 16. Lim_n→∞{1³ + 2³ + ................ + n³}/(n⁴)

Solution:

We have,

Lim_n→∞{1³ + 2³ + ................ + n³}/(n⁴)

= Lim_n→∞[n²(n + 1)²]/(4n⁴)           [since (1³ + 2³ + ............. + n³) = n²(n + 1)²/4]

= Lim_n→∞[(n + 1)²]/(4n²)

= Lim_n→∞[(1 + 1/n)² × (1/4)]

When n → ∞, (1/n) → 0

= 1/4

Question 17.  Lim_n→∞{1³ + 2³ + ................ + n³}/(n - 1)⁴

Solution:

We have,

Lim_n→∞{1³ + 2³ + ................ + n³}/(n - 1)⁴

= Lim_n→∞[n²(n + 1)²]/[4(n - 1)⁴]           [since (1³ + 2³ + ............. + n³) = n²(n + 1)²/4]

=

=

When n → ∞, (1/n) → 0

= 1/4

Question 18. Lim_x→∞[√x{√(x + 1) - √x}]

Solution:

We have,

Lim_x→∞[√x{√(x + 1) - √x}]

On rationalizing numerator, we get

= Lim_x→∞[(√x){(x + 1) - x}]/{√(x + 1) + √x}

= Lim_x→∞(√x}/{√(x + 1) + √x}

=

When x → ∞, (1/x) → 0.

= 1/(√1 + 1)

= 1/2

Question 19. Lim_x→∞[1/3 + 1/3² + 1/3³ + .................. + 1/3ⁿ]

Solution:

We have,

 Lim_x→∞[1/3 + 1/3² + 1/3³ + .................. + 1/3ⁿ]

This is G.P series of common ratio 1/3.

So, the sum of n terms of G.P. S_n = [a(1 - rⁿ)]/(1 - r)        (i)

a = 1/3, r = 1/3

On putting the value of a & r in equation (i), we get

S_n = (1/2)(1 - 1/3ⁿ) 

= Lim_x→∞[(1/2)(1 - 1/3ⁿ)]

= (1/2)Lim_x→∞(1 - 1/3ⁿ)

= (1/2)(1 - 0)

= 1/2

Question 20. Lim_x→∞{(x⁴ + 7x³ + 46x + a)}/{(x⁴ + 6)}.

Solution:

We have,

 Lim_x→∞{(x⁴ + 7x³ + 46x + a)}/{(x⁴ + 6)}

=

When x → ∞, (1/x), (1/x²), (1/x³), (1/x⁴) → 0

= 1/1

= 1

Question 21. f(x) = (ax² + b)/(x² + 1), Lim_x→0f(x) = 1, Lim_x→∞f(x) = 1, then prove that f(-2) = f(2) = 1

Solution:

We have,

 f(x) = (ax² + b)/(x² + 1)

= Lim_x→0[(ax² + b)/(x² + 1)]

b/1 = 1

b = 1

= Lim_x→∞[(ax² + b)/(x² + 1)]

= 

When x → ∞, (1/x²) → 0.

(a + 0)/(1 + 0) = 1

a = 1

Hence, a = 1, b = 1

f(x) = (x² + 1)/(x² + 1)

f(x) = 1

f(-2) = 1

f(2) = 1   (Since f(x) is independent on x)

f(-2) = f(2) = 1

Hence proved

Question 22. Show that Lim_x→∞[√(x² + x + 1) - x] ≠ Lim_x→∞[√(x² + 1) - x]

Solution:

We have,

L.H.S,

= Lim_x→∞[√(x² + x + 1) - x]

On rationalizing numerator, we get

= Lim_x→∞[(x² + x + 1) - x²]/[√(x² + x + 1) + x]

= Lim_x→∞(x + 1)/[√(x² + x + 1) + x]

= Lim_x→∞[x(1 + 1/x)/[x{√(1 + 1/x + 1/x²) + 1}]

= Lim_x→∞[(1 + 1/x)/[{√(1 + 1/x + 1/x²) + 1}]

When x → ∞, (1/x), (1/x²) → 0.

= 1/(√1 + 1)

= 1/2

Now we solve R.H.S,

= Lim_x→∞[√(x² + 1) - x]

On rationalizing numerator, we get

= Lim_x→∞[(x² + 1) - x²]/[√(x² + 1) + x]

= Lim_x→∞(1)/[√(x² + 1) + x]

= 1/[√(∞ + 1) + ∞]

= 1/∞

= 0

L.H.S ≠ R.H.S

Hence, Lim_x→∞[√(x² + x + 1) - x] ≠ Lim_x→∞[√(x² + 1) - x]

Question 23. Lim_x→-∞[√(4x² - 7x) + 2x]

Solution:

We have,

 Lim_x→-∞[√(4x² - 7x) + 2x]

Let x = -n when x → -∞, then n → ∞.

= Lim_n→∞[√(4n² + 7n) - 2n]

On rationalizing numerator, we get

= Lim_n→∞[(4n² + 7n) - 4n²]/[√(4n² + 7n) + 2n]

= Lim_n→∞[(7n)/[√(4n² + 7n) + 2n]

= Lim_n→∞(7n)/[n{√(4 + 7/n) + 2}]

= Lim_n→∞(7)/{√(4 + 7/n) + 2}

When n → ∞, (1/n) → 0

= 7/(√4 + 2)

= 7/(2 + 2)

= 7/4

Question 24. Lim_x→-∞[√(x² - 8x) + x]

Solution:

We have,

Lim_x→-∞[√(x² - 8x) + x]

Let x = -n when x → -∞, then n → ∞.

= Lim_n→∞[√(n² + 8n) - n]

On rationalizing numerator, we get

= Lim_n→∞[(n² + 8n) - n²]/[√(n² + 8n) + n]

= Lim_n→∞[(8n)/[√(n² + 8n) + n]

= Lim_n→∞(8n)/[n{√(1 + 8/n) + 1}]

= Lim_n→∞(8)/{√(1 + 8/n) + 1}

When n → ∞, (1/n) → 0

= 8/(√1 + 1)

= 8/2

= 4

Question 25. Lim_n→∞(1⁴ + 2⁴ + ..........+ n⁴)/n⁵ - Lim_n→∞(1³ + 2³ + .......... + n³)/n⁵

Solution:

We have,

Lim_n→∞(1⁴ + 2⁴ + ..........+ n⁴)/n⁵ - Lim_n→∞(1³ + 2³ + .......... + n³)/n⁵

=

=

=

=

When n → ∞, (1/n), (1/n²), (1/n³) → 0

= 1/3 × 1 × 2 × 3 - 1/4 × 0

= 6/30

= 1/5

Question 26. Lim_n→∞{(1.2 + 2.3 + 3.4 + ..........+ n (n + 1)}/n³

Solution:

We have,

Lim_n→∞{(1.2 + 2.3 + 3.4 + ..........+ n (n + 1)}/n³

=

=

=

=

=

=

When n → ∞, (1/n) → 0

= (1 × 2)/6

= 2/6

= 1/3

Summary

Exercise 29.6 Set 2 covers the evaluation of limits of algebraic functions, including polynomials and rational functions, as x approaches infinity or 0. This is a crucial concept in calculus, as it helps to understand the behavior of functions and their properties. The exercise includes a variety of questions that require the application of different limit properties and theorems, such as the sum, product, and chain rule, as well as the squeeze theorem. Students are expected to use algebraic manipulations, such as factoring and canceling common factors, to evaluate the limits. The exercise also includes questions that require the use of L'Hopital's rule, which is a powerful tool for evaluating limits of functions that approach 0/0 or ∞/∞. By working through these questions, students will develop a deeper understanding of limits and their applications in calculus, as well as improve their problem-solving skills and ability to think critically.